Solutions to the equation $ze^{a-z}=1$

Solution 1:

I may have the solution. On the unit disk, by the Maximum Modulus Theorem, we have: $$\left|ze^{a-z}-1-(ze^{a-z})\right|=1\leq\left|ze^{a-z}\right|=e^t, t>0$$

So by Rouche's Theorem, $ze^{a-z}-1$ has the same number of zeros as $ze^{a-z}$ in the unit disk. Since $e^{a-z}$ has no zeroes, $ze^{a-z}$ has one zero (at $z=0$ ) in the unit disk. Thus $$ze^{a-z}=1$$ has one solution in the unit disk. If $z$ is real, we have the equation $$e^x=e^ax, a>1$$

This must have a solution, because $e^a(0)=0<e^0=1$ but $e^a(1)=e^a>e^1=e.$ Since both functions are continuous, they must have a real solution somewhere in between $0<x<1$.

So the answer is that there is exactly one solution, and it is a real solution.